Evaluate $\int\limits_{0}^{2\pi} |\sin x| dx$.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
Let $I = \int\limits_{0}^{2\pi} |\sin x| dx$
$= \int\limits_{0}^{\pi} |\sin x| dx + \int\limits_{\pi}^{2\pi} |\sin x| dx$
$= \int\limits_{0}^{\pi} \sin x \, dx - \int\limits_{\pi}^{2\pi} \sin x \, dx$
$= [-\cos x]_{0}^{\pi} - [-\cos x]_{\pi}^{2\pi}$
$= [-\cos \pi + \cos 0] - [-\cos 2\pi + \cos \pi]$
$= [1 + 1] - [-1 - 1] = 2 + 2 = 4$